AtCoder Beginner Contest 127
F
Absolute Minima
AtCoder
⌀ 2003
原题
← E→
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Problem Statement

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There is a function $f(x)$, which is initially a constant function $f(x) = 0$.

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We will ask you to process $Q$ queries in order. There are two kinds of queries, update queries and evaluation queries, as follows:

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  • An update query 1 a b: Given two integers $a$ and $b$, let $g(x) = f(x) + |x - a| + b$ and replace $f(x)$ with $g(x)$.
  • An evaluation query 2: Print $x$ that minimizes $f(x)$, and the minimum value of $f(x)$. If there are multiple such values of $x$, choose the minimum such value.
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We can show that the values to be output in an evaluation query are always integers, so we ask you to print those values as integers without decimal points.

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Constraints

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  • All values in input are integers.
  • $1 \leq Q \leq 2 \times 10^5$
  • $-10^9 \leq a, b \leq 10^9$
  • The first query is an update query.
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Input

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Input is given from Standard Input in the following format:

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Q
Query_1
:
Query_Q

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See Sample Input 1 for an example.

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Output

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For each evaluation query, print a line containing the response, in the order in which the queries are given.

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The response to each evaluation query should be the minimum value of $x$ that minimizes $f(x)$, and the minimum value of $f(x)$, in this order, with space in between.

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样例

Input #1
4
1 4 2
2
1 1 -8
2
Output #1
4 2
1 -3
In the first evaluation query, $f(x) = |x - 4| + 2$, which attains the minimum value of $2$ at $x = 4$. In the second evaluation query, $f(x) = |x - 1| + |x - 4| - 6$, which attains the minimum value of $-3$ when $1 \leq x \leq 4$. Among the multiple values of $x$ that minimize $f(x)$, we ask you to print the minimum, that is, $1$.
Input #2
4
1 -1000000000 1000000000
1 -1000000000 1000000000
1 -1000000000 1000000000
2
Output #2
-1000000000 3000000000
题目信息
题号
abc127_f
时间限制
2 s
内存限制
1024 MB

题面与样例版权归 AtCoder 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。